Take any triangle ABC. Construct equilateral triangles externally on each side. Let A', B' and C' be the external vertices of the externally constructed equilateral triangles with A' off the side opposite vetrex A, etc.

Construct Lines AA', BB' and CC'. Observe. Are they concurrent? Is the point of concurrency any of the orthocenter, centroid, incenter or circumcenter of triangle ABC? If not what can you find out about this point.

Based on our earlier discussion we know that the point of intersection is the Fermat point. The Fermat point is none is the orthocenter, centroid, incenter or circumcenter of triangle ABC - certainly we can alter triangle ABC so that the Fermat point co-incides with one or more of the centers (consider the trivial case where ABC is equilateral and the Fermat point is all of the above) but it will not be any of these centers at all times.

Some important remarks regarding the Fermat point include:

1. The Fermat point will only be interior to triangle ABC provided that no angle of triangle ABC exceeds 120deg. If any angle equals 120deg then the Fermat point is concurrent with the vertex of that angle and should any angle exceed 120deg the Fermat point will be exterior to triangle ABC, click here to see a GSP animation of these remarks.

2. Line segments AA', BB' and CC' are equal in length (each being equal to AF+ BF + CF) and meet each other at 60deg.

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